6.1 Power Series and Functions

Big idea

A polynomial is the friendliest function there is. You evaluate it with nothing but addition and multiplication, you differentiate and integrate it by a rule you can carry out in your head, and its behaviour holds no surprises. Almost every function worth studying is not a polynomial. A power series is the attempt to keep the convenience anyway: an expression built entirely from powers of $x$ and constant coefficients, but with infinitely many terms, and therefore with enough room to be a sine, a logarithm, or the solution of a differential equation that has no name.

The price of infinitely many terms is that the expression does not automatically mean anything. Once you fix a value of $x$, however, the ambiguity disappears. The series stops being a formula in a variable and becomes an ordinary numerical series, and deciding whether it converges is a question you already have tools for. So a power series is really a whole family of numerical series indexed by $x$, and the first job is to sort the values of $x$ into the ones where the family member converges and the ones where it does not.

That sorting turns out to be far tidier than it has any right to be. The set of good $x$ values is never scattered. It is always an interval, and always centred at the same point the series is built around. There is a single number $R$, the radius of convergence, that measures how far you may move from the centre before the terms stop shrinking fast enough. Inside that distance you get convergence, and convergence of the strongest kind; outside it you get divergence, with the terms failing even to approach zero.

What is left over is the boundary. At exactly distance $R$ from the centre the ratio test goes silent, and the two endpoints have to be examined one at a time by other means. This is not a technicality you can skip. The endpoint behaviour is what distinguishes the four possible shapes of the answer, and it is where most of the graded errors on this topic live.

Decoder

For a power series centred at a point $a$, exactly one of three things happens: it converges only when $x = a$, it converges for every real $x$, or there is a positive number $R$ with convergence whenever $|x - a| < R$ and divergence whenever $|x - a| > R$.

The word carrying the content is exactly. There is no fourth possibility, and in particular there is no power series that converges on a lopsided set or on a set full of holes. Once you know $R$, you know the answer everywhere except at the two points $x = a - R$ and $x = a + R$, and the statement deliberately says nothing about those.

The first case is $R = 0$ and the second is $R = \infty$, so the three cases are one statement with the two degenerate values allowed. Notice that $x = a$ always works: every term past the first is a power of zero, so the series collapses to its constant term. A power series can never fail to converge at its own centre, which is why “converges only at $a$” is the smallest possible answer rather than “converges nowhere”.

The classic mistake is to report the interval as soon as the ratio test produces an inequality. The ratio test hands you a strict inequality $|x - a| < R$, and a strict inequality has open ends. Whether each end belongs is a separate question with a separate answer, decided by substituting that single number and running a convergence test on the resulting numerical series. Do it twice, once per endpoint, and do not assume the two ends behave alike.

Definitions and results

Power series. A power series centred at $a$ is an expression

$$ \sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + \cdots $$

where the $c_n$ are fixed numbers called the coefficients. Taking $a = 0$ gives the common special case $\sum c_n x^n$.

Convergence is pointwise. To say the series converges at a particular $x$ means the sequence of partial sums of that numerical series has a finite limit. Different $x$ values are independent questions, at least before the theory below ties them together.

The three cases. Exactly one of the following holds: the series converges only at $x = a$; it converges for all real $x$; or there is $R > 0$ such that it converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$. The number $R$ is the radius of convergence, with $R = 0$ and $R = \infty$ used for the first two cases.

Interval of convergence. The set of all $x$ where the series converges. When $R$ is finite and positive it is one of $(a-R, a+R)$, $[a-R, a+R)$, $(a-R, a+R]$ or $[a-R, a+R]$. When $R = \infty$ it is all of $(-\infty, \infty)$, and when $R = 0$ it is the single point $a$.

Finding $R$ by the ratio test. Apply the ratio test to the absolute values of the terms and read off the limit:

$$ \lim_{n \to \infty} \left| \frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n} \right| = |x - a| \cdot \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| $$

Convergence is guaranteed when this limit is less than $1$ and divergence when it is greater than $1$, so $R$ is the reciprocal of the coefficient limit whenever that limit exists and is nonzero.

The ratio test says nothing at the endpoints. At $x = a \pm R$ the limit above equals $1$ exactly, which is the inconclusive case. Substitute the endpoint value and test the resulting constant series directly, typically with the alternating series test, a $p$-series comparison, or the divergence test.

Absolute convergence inside. Strictly inside the interval the convergence is absolute, so you may rearrange terms freely and the series behaves like a finite sum for most purposes. At an endpoint the convergence may be only conditional.

A power series defines a function. On its interval of convergence, the assignment $x \mapsto \sum c_n(x-a)^n$ is a genuine function with that interval as its domain. Questions about the function are answered through the series, which is why the domain question comes first.

Worked examples

A geometric series in disguise

Consider $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{2^n}$. Each term is $(x/2)^n$, so this is a geometric series with ratio $x/2$. Geometric series converge exactly when the ratio has absolute value below $1$, which means $|x| < 2$, and the radius of convergence is $2$.

The endpoints are $x = 2$ and $x = -2$. At $x = 2$ the terms are all $1$ and the partial sums march off to infinity. At $x = -2$ the terms alternate between $1$ and $-1$ and never approach zero. Both diverge, so the interval of convergence is $(-2, 2)$, open at both ends.

Inside that interval the sum is known:

$$ \sum_{n=0}^{\infty} \left( \frac{x}{2} \right)^{n} = \frac{1}{1 - x/2} = \frac{2}{2 - x} $$

Check at $x = 1$: the series is $1 + \tfrac12 + \tfrac14 + \cdots = 2$, and the formula gives $2/(2-1) = 2$.

Endpoints that disagree

Consider $\displaystyle \sum_{n=1}^{\infty} \frac{(x-2)^n}{n \, 3^n}$, centred at $a = 2$. The ratio of consecutive absolute terms is

$$ \frac{|x-2|^{n+1}}{(n+1)3^{n+1}} \cdot \frac{n \, 3^n}{|x-2|^n} = \frac{|x-2|}{3} \cdot \frac{n}{n+1} $$

and the second factor tends to $1$, so the limit is $|x-2|/3$. That is below $1$ precisely when $|x - 2| < 3$, giving $R = 3$ and the open interval $-1 < x < 5$.

Now test the ends. At $x = 5$ the numerator becomes $3^n$, which cancels the $3^n$ below and leaves $\sum 1/n$, the harmonic series, which diverges. At $x = -1$ the numerator becomes $(-3)^n$ and the series reduces to $\sum (-1)^n/n$, which converges by the alternating series test since $1/n$ decreases to zero. The two ends genuinely behave differently, and the interval of convergence is $[-1, 5)$.

A radius of zero

Consider $\displaystyle \sum_{n=0}^{\infty} n!\, x^n$. The ratio of absolute terms is

$$ \frac{(n+1)!\,|x|^{n+1}}{n!\,|x|^{n}} = (n+1)|x| $$

For any fixed $x \neq 0$ this grows without bound, so the ratio test reports divergence. Only $x = 0$ survives, where the series is just its constant term $0! \cdot 1 = 1$. Here $R = 0$ and the interval of convergence is the single point $\{0\}$. Factorials in the numerator overwhelm any fixed power.

A radius of infinity

Consider $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$, the same factorials moved downstairs. The ratio is

$$ \frac{|x|^{n+1}}{(n+1)!} \cdot \frac{n!}{|x|^{n}} = \frac{|x|}{n+1} $$

which tends to $0$ for every $x$, no matter how large. The limit is below $1$ always, so $R = \infty$ and the interval of convergence is all of $(-\infty, \infty)$. There are no endpoints to test, which is the one case where skipping that step is correct.

Practice

Work the two halves of the question separately every time: first get the radius from the ratio test, then substitute each endpoint and decide it on its own. These drills ask for the radius, the interval, or both.

Quiz

Five items on radii of convergence, endpoint testing, and reading off the interval.